LC Higher Level Solutions Sample Paper 2
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Sample Paper 2: Paper 1
Question 2 (25 marks)
Question 2 (a)
z = 2 + 2 3i
z = 22 + (2 3 ) 2 = 4 + 12 = 16 = 4
z = r (cos θ + i sin θ )
2 3
π
= 3 ⇒ θ = tan −1 ( 3 ) = 60o =
2
3
π
π

∴ z = 4  cos + i sin 
3
3

tan θ =
z = r , arg z = θ
Im
4
z
w = 3 +i
3
w = ( 3 ) 2 + (1) 2 = 3 + 1 = 4 = 2
2
1
π
 1 
o
⇒ θ = tan −1 
 = 30 = 6
3
 3
π
π

∴ w = 2  cos + i sin 
6
6

tan θ =
1
0
w
60o 30o
1
Re
2
3
4
Question 2 (b) (i)
Question 2 (b) (ii)
π
π 
π
π

zw = 4  cos + i sin  2  cos + i sin 
3
3 
6
6

π
π

= 8  cos + i sin 
2
2

= 8(0 + i )
= 8i
‘According to De Moivre’s theorem, when
two complex numbers are multiplied together,
the arguments are added and the moduli are
multiplied to form the new complex number.’
Question 2 (c) (i)
Question 2 (c) (ii)
π
π

4  cos + i sin 
z
π
π
3
3

= 
= 2  cos + i sin 
π
π
w
6
6


2  cos + i sin 
6
6

 3 1 
= 2 
+ i  = 3 + i
2
2 

‘According to De Moivre’s theorem, when two
complex numbers are divided, the arguments
are subtracted and the moduli are divided to
form the new complex number.’
Question 2 (c) (iii)
π
π
z
n
n
14 
  = 2  cos + i sin  [r (cos θ + i sin θ )] = r (cos nθ + i sin nθ ) (De Moivre’s Theorem)
6
6
 w

14π
14π 

= 214  cos
+ i sin

6
6 

14
14
7π
7π  14  1
3  13

= 214  cos
+ i sin
i  = 2 (1 + 3i )
 = 2  +
3
3 

2 2 